Using tangent substitution on $Bbb R$
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How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
$endgroup$
add a comment |
$begingroup$
How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
$endgroup$
1
$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15
add a comment |
$begingroup$
How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
$endgroup$
How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
calculus integration
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
edited Dec 2 '18 at 23:09
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Dec 2 '18 at 20:09
GaboruGaboru
637
asked Dec 2 '18 at 20:09
GaboruGaboru
637
asked Dec 2 '18 at 20:09
GaboruGaboru
637
637
1
$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15
add a comment |
1
$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15
1
1
$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15
add a comment |
1 Answer
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$begingroup$
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
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add a comment |
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$begingroup$
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
$endgroup$
add a comment |
$begingroup$
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
$endgroup$
add a comment |
$begingroup$
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
$endgroup$
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
edited Dec 9 '18 at 19:45
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
answered Dec 2 '18 at 23:07
Jean MarieJean Marie
29.3k42050
29.3k42050
add a comment |
add a comment |
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$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20
$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41
$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15